Question

The line parallel to the x-axis and passing through the point of intersection of the lines $ax+2by+3b=0$ and $bx-2ay-3a=0,$ where $(a,b)\ne (0,0)$ is

• A. above the x-axis at a distance of $\frac{3}{2}$
• B. above the x-axis at a distance of $\frac{2}{3}$
• C. below the x-axis at a distance of $\frac{2}{3}$
• D. below the x-axis at a distance of $\frac{3}{2}$
• E. below the x-axis at a distance of 3

Answer 1

Answer: (D)

Let the line parallel to $x-$ axis is, $y=c$ ...(i) Given lines, $ax+2by=-3b$ ...(ii) $bx-2ay=3a$ ...(iii) Multiply by b in Eq. (ii) and by a in Eq. (iii), then subtract Eq. (ii) from Eq. (iii), $\begin{array}{rl}& \underline{\begin{array}{rl}& abx+2b^{2}y=-3b^2\\ & xab-2a^{2}y=3a^2\\ & -+-\end{array}}\\ & 2(a^2+b^2)y=-3(a^2+b^2)\end{array}$
$\Rightarrow$ $y=-\frac{3}{2}$ From Eq. (i), $c=-\frac{3}{2}$ Hence, the line is, $y=-\frac{3}{2}$ Which is passing through the point of intersection of the lines Eqs. (ii) and (iii) below the $x-$ axis at a distance of $\frac{3}{2}$ .